Velocity Question height, maximum height A projectile is fired upward from a height 200 ft, and reaches a maximum height of 400 ft.
My Approach
Using this equation:
$$s(t)=s_0+v_0t+\frac{1}{2}gt^2$$
I plug in my knowns and get this following equation:
$$400=200+v_0t+16t^2$$
I reduce it down and get this equation:
$$200=v_0t+16t^2$$
How could I get the initial velocity? I know from here that this is when t is at the maximum height? Only using position, and velocity equations
 A: $$v^2=u^2+2as$$
$$0^2=u^2+2(-g)(200)$$
$$u^2=400g$$
$$u=20\sqrt{g}\approx62.6 fts^{-1}$$
A: Initial velocity is the same as a final velocity of a free falling from $400\ \mathrm{ft.}$ to $200\ \mathrm{ft.}$, i. e. from $200\ \mathrm{ft.}$ to the ground, so the equation is as simple as
$$s = {1\over 2}gt^2$$
where $s = 200\ \mathrm{ft.} = 200\times0.3048\ \mathrm m = 60.96\ \mathrm m \ $ and $\ \ g\approx 9.8\ {\mathrm m \mathrm s^{-2} }$. It means that
$$t = \sqrt{{2s \over g}}$$
and the initial velocity is
$$v = gt = g\sqrt{{2s \over g}}= \sqrt{2sg} \approx  \sqrt{2\times 60.96 \times 9.8}\ {\mathrm m \over \mathrm s} \approx 34.6\ {\mathrm m \over \mathrm s}\cdot$$
A: If you insist on deriving all from the formulas
\begin{align}
v(t) &= v_0 + at \tag 1\\
s(t) &=s_0+v_0t+\frac{1}{2}at^2\tag 2
\end{align}
only, let you will be fulfilled.
(Note: $(2)$ is derived from $(1)$ by integration.)
First, the acceleration is negative, as the Earth attraction works against the movement, so $a = -g,\ $ and $(1)$ and $(2)$ become
\begin{align}
v(t) &= v_0 - gt \tag 3\\
s(t) &=s_0+v_0t-\frac{1}{2}gt^2\tag 4
\end{align}
Now, to better understand the situation, substitute in $(3)$ known values $s_0 = 200\ \mathrm {ft}\,$ and $g \approx 32.17\ {\mathrm {ft}\cdot \mathrm s}^{-2},\ $ and (meanwhile) unknown value $v_0 \approx 113.4\ \mathrm{ft}\cdot {\mathrm s}^{-1}\ $ (which we will calculate at the end).
We get this graph - a linear dependency of a speed from time:

We may see that the speed of the projectile decreases from the speed $v_0 \approx 113.4\ \mathrm{ft}\cdot {\mathrm s}^{-1}\ \ $ (in time $t_0=0)\ $ to zero speed $v_1 = 0\ $ (in time $t_1 \approx 3.5\ \mathrm s)$.
Substituting the same values into $(4)$ we obtain this graph - a quadratic dependency of the position of the projectile from time:

We may see that the position of the projectile increases from the position $s_0 = 200\ \mathrm{ft}\ \ $ (in time $t_0=0)\ $ to the position $s_1 = 400\ \mathrm{ft}\ \ $ (in time $t_1 \approx 3.5\ \mathrm s)$.
So let $t_1$ is the time in which the projectile reaches its maximum height of $s_1 = s(t_1) \ $ (and we know that $s_1 = 400\ \mathrm{ft}.$) Equations $(3)$ and $(4)$ become
\begin{align}
v(t_1) &= v_0 - gt_1 \tag 5\\
s(t_1) &= s_0+v_0t_1-\frac{1}{2}gt_1^2\tag 6
\end{align}
But  $\ v(t_1) = 0\ $ and $\ s(t_1) = s_1$:
\begin{align}
0 &= v_0 - gt_1 \tag 7\\
s_1 &= s_0+v_0t_1-\frac{1}{2}gt_1^2\tag 8
\end{align}
From $(7)$ we obtain
$$v_0 = gt_1\tag 9$$
and substituting it into $(8)$ we obtain
$$s_1 =s_0+g{t_1}^2-\frac{1}{2}gt_1^2$$
i. e.
$$s_1 =s_0+\frac{1}{2}gt_1^2 \tag {10}$$
Now it is all known, except $t_1$, so let express it from $(10)$:
$$t_1 = \sqrt{{2(s_1-s_0)\over g}}$$
Substituting it into $(9)$ gives us the result
$$v_0 = gt_1 = g\sqrt{{2(s_1-s_0)\over g}} = \sqrt{{2(s_1-s_0) g}}$$
in particular
$$v_0  = \sqrt{2\times 200 \times 32,17}\ {\mathrm {ft}\over\mathrm m} \approx 113.4 {\text{ft}\over\text{m}}$$
A: By the conservation of energy we get that $$\frac{1}{2}mv^2=mgh\implies v=\sqrt{2gh}=\sqrt{400g}=20\sqrt{32.2}\text{ft/s}$$
